Water flows at 0.65 m/s through a 3.0 cm diameter hose that terminates in a 0.3 cm diameter nozzle. (Recall, the density of water is 1000 kg/m3.) (a) What is the speed of the water leaving the nozzle

v2 = 65 m/s

the speed of the water leaving the nozzle is 65 m/s

Explanation:

Given;

Water flows at 0.65 m/s through a 3.0 cm diameter hose that terminates in a 0.3 cm diameter nozzle

Initial speed v1 = 0.65 m/s

diameter d1 = 3.0 cm

diameter (nozzle) d2 = 0.3 cm

The volumetric flow rates in both the hose and the nozzle are the same.

V1 = V2 1

Volumetric flow rate V = cross sectional area × speed of flow

V = Av

Area = (πd^2)/4

V = v(πd^2)/4 2

Substituting equation 2 to 1;

v1(πd1^2)/4 = v2(πd2^2)/4

v1d1^2 = v2d2^2

v2 = (v1d1^2)/d2^2

Substituting the given values;

v2 = (0.65 × 3^2)/0.3^2

v2 = 65 m/s

the speed of the water leaving the nozzle is 65 m/s

v=ir = 3x15 =45 v

find the surface area of the sphere to the nearest square unit. use a calculator.

a. 254 in.2

b. 64 in.2

c. 127 in.2

d. 1,018 in.2